Question: Divide the following complex numbers. $ \dfrac{-7-19i}{5-4i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5+4i}$ $ \dfrac{-7-19i}{5-4i} = \dfrac{-7-19i}{5-4i} \cdot \dfrac{{5+4i}}{{5+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-7-19i) \cdot (5+4i)} {(5-4i) \cdot (5+4i)} = \dfrac{(-7-19i) \cdot (5+4i)} {5^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-7-19i) \cdot (5+4i)} {(5)^2 - (-4i)^2} = $ $ \dfrac{(-7-19i) \cdot (5+4i)} {25 + 16} = $ $ \dfrac{(-7-19i) \cdot (5+4i)} {41} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-7-19i}) \cdot ({5+4i})} {41} = $ $ \dfrac{{-7} \cdot {5} + {-19} \cdot {5 i} + {-7} \cdot {4 i} + {-19} \cdot {4 i^2}} {41} $ Evaluate each product of two numbers. $ \dfrac{-35 - 95i - 28i - 76 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{-35 - 95i - 28i + 76} {41} = \dfrac{41 - 123i} {41} = 1-3i $